Optimal. Leaf size=205 \[ \frac {(a+b) \log \left (\sqrt [3]{k x^3+(-k-1) x^2+x}-\sqrt [3]{b} x\right )}{b^{2/3}}+\frac {(-a-b) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 b^{2/3}}+\frac {\left (\sqrt {3} a+\sqrt {3} b\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{b^{2/3}}+\frac {3 \sqrt [3]{k x^3+(-k-1) x^2+x}}{x} \]
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Rubi [F] time = 25.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{(1-x)^{2/3} x^{5/3} (1-k x)^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \left (-\frac {2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )}{(b-k)^2 (1-x)^{2/3} x^{5/3} (1-k x)^{2/3}}-\frac {(1+k) (a+k)}{(b-k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}}-\frac {(a+b) \left (1+2 b+k^2\right )-(a+b) (1+k) \left (1+3 b-k+k^2\right ) x}{(b-k)^2 (1-x)^{2/3} x^{5/3} (1-k x)^{2/3} \left (1+(-1-k) x+(-b+k) x^2\right )}\right ) \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=-\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {(a+b) \left (1+2 b+k^2\right )-(a+b) (1+k) \left (1+3 b-k+k^2\right ) x}{(1-x)^{2/3} x^{5/3} (1-k x)^{2/3} \left (1+(-1-k) x+(-b+k) x^2\right )} \, dx}{(b-k)^2 ((1-x) x (1-k x))^{2/3}}-\frac {\left ((1+k) (a+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {1}{(1-x)^{2/3} x^{2/3} (1-k x)^{2/3}} \, dx}{(b-k) ((1-x) x (1-k x))^{2/3}}-\frac {\left (\left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {1}{(1-x)^{2/3} x^{5/3} (1-k x)^{2/3}} \, dx}{(b-k)^2 ((1-x) x (1-k x))^{2/3}}\\ &=\frac {3 \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x) (1-k x)}{2 (b-k)^2 ((1-x) x (1-k x))^{2/3}}+\frac {3 (1+k) (a+k) (1-x) \left (\frac {(1-k) x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {1-x}{1-k x}\right )}{(1-k) (b-k) ((1-x) x (1-k x))^{2/3}}-\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \left (\frac {-\left ((a+b) (1+k) \left (1+3 b-k+k^2\right )\right )-\frac {(a+b) \left (1+5 b+4 b^2-k+2 b k+5 b k^2-k^3+k^4\right )}{\sqrt {1+4 b-2 k+k^2}}}{(1-x)^{2/3} x^{5/3} (1-k x)^{2/3} \left (-1-k-\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )}+\frac {-\left ((a+b) (1+k) \left (1+3 b-k+k^2\right )\right )+\frac {(a+b) \left (1+5 b+4 b^2-k+2 b k+5 b k^2-k^3+k^4\right )}{\sqrt {1+4 b-2 k+k^2}}}{(1-x)^{2/3} x^{5/3} (1-k x)^{2/3} \left (-1-k+\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )}\right ) \, dx}{(b-k)^2 ((1-x) x (1-k x))^{2/3}}+\frac {\left ((-1-k) \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {1}{(1-x)^{2/3} x^{2/3} (1-k x)^{2/3}} \, dx}{2 (b-k)^2 ((1-x) x (1-k x))^{2/3}}\\ &=\frac {3 \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x) (1-k x)}{2 (b-k)^2 ((1-x) x (1-k x))^{2/3}}+\frac {3 (1+k) (a+k) (1-x) \left (\frac {(1-k) x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {1-x}{1-k x}\right )}{(1-k) (b-k) ((1-x) x (1-k x))^{2/3}}+\frac {3 (1+k) \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x) \left (\frac {(1-k) x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {1-x}{1-k x}\right )}{2 (1-k) (b-k)^2 ((1-x) x (1-k x))^{2/3}}-\frac {\left (\left (-\left ((a+b) (1+k) \left (1+3 b-k+k^2\right )\right )+\frac {(a+b) \left (1+5 b+4 b^2-k+2 b k+5 b k^2-k^3+k^4\right )}{\sqrt {1+4 b-2 k+k^2}}\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {1}{(1-x)^{2/3} x^{5/3} (1-k x)^{2/3} \left (-1-k+\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )} \, dx}{(b-k)^2 ((1-x) x (1-k x))^{2/3}}+\frac {\left ((a+b) \left (1+k^3+3 b (1+k)+\frac {4 b^2+(1-k)^2 \left (1+k+k^2\right )+b \left (5+2 k+5 k^2\right )}{\sqrt {4 b+(-1+k)^2}}\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {1}{(1-x)^{2/3} x^{5/3} (1-k x)^{2/3} \left (-1-k-\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )} \, dx}{(b-k)^2 ((1-x) x (1-k x))^{2/3}}\\ \end {align*}
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Mathematica [F] time = 14.01, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 0.42, size = 205, normalized size = 1.00 \begin {gather*} \frac {3 \sqrt [3]{x+(-1-k) x^2+k x^3}}{x}+\frac {\left (\sqrt {3} a+\sqrt {3} b\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {(a+b) \log \left (-\sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}+\frac {(-a-b) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \end {gather*}
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b - k\right )} x^{2} + {\left (k + 1\right )} x - 1\right )} x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (-2+\left (1+k \right ) x \right ) \left (1-\left (1+k \right ) x +\left (a +k \right ) x^{2}\right )}{x \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-\left (1+k \right ) x +\left (-b +k \right ) x^{2}\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b - k\right )} x^{2} + {\left (k + 1\right )} x - 1\right )} x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {\left (x\,\left (k+1\right )-2\right )\,\left (\left (a+k\right )\,x^2+\left (-k-1\right )\,x+1\right )}{x\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b-k\right )\,x^2+\left (k+1\right )\,x-1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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